Convergence of Discrete-Time Deterministic Games to Path-Dependent Isaacs Partial Differential Equations Under Quadratic Growth Conditions

Abstract

We consider discrete-time approximations for path-dependent Isaacs partial differential equations (PDEs) of deterministic differential games under quadratic growth conditions including linear/quadratic problems with distributed and discrete delays. Owing to the path-dependence of the system, the Isaacs PDEs are defined on infinite-dimensional spaces of past state trajectories. Using the notion of viscosity solutions on the infinite-dimensional spaces as proposed by Lukoyanov based on co-invariant derivatives of path spaces, we show that the discrete-time path-dependent dynamic games converge to a unique viscosity solution for the Isaacs PDEs. Noting that these games can be practically defined on finite-dimensional spaces, we discuss finite-dimensional approximations of viscosity solutions of path-dependent Isaacs PDEs. Given an example, we derive discrete-time Riccati-type recursive equations to calculate explicit discrete-time approximations for the path-dependent linear/quadratic problems.

Proofs of Uniform Estimates on \(\{ V^{(N)} \}_{N=1}^\infty \)

Proof of Proposition 3.3

Let $t_i \le t<t_{i+1}$ and $x_t \in {\mathbf {X}}_{t}$. Suppose there exists ${\hat{K}}_{i+1} \ge 0$ such that

$$\begin{aligned} V^{(N)}(t_{i+1},y_{t_{i+1}}) \ge -{\hat{K}}_{i+1}, \quad \forall y_{t_{i+1}} \in {\mathbf {X}}_{t_{i+1}}. \end{aligned}$$

By (3.2) with (A3),

$$\begin{aligned} V^{(N)}(t,x_t)&\ge -K_l \varDelta _N-{\hat{K}}_{i+1}=:-{\hat{K}}_{i}. \end{aligned}$$

Noting that $V^{(N)}(T,x_T) = \varPhi (x_T) \ge -K_\varPhi$, we have the recursive equation of ${\hat{K}}_i$:

$$\begin{aligned}&{\hat{K}}_N=K_\varPhi , \\&{\hat{K}}_i=K_l \varDelta _N +{\hat{K}}_{i+1} \quad (i=N-1,N-2,\ldots , 0). \end{aligned}$$

Solving the above recursive equations, we have ${\hat{K}}_0=K_\varPhi +K_lT$. Hence we obtain

$$\begin{aligned} {\hat{K}}_i \le {\hat{K}}_0=K_\varPhi +K_lT \quad (i=0,1,\ldots , N). \end{aligned}$$

Let $t_i \le t <t_{i+1}$ and $x_t \in {\mathbf {X}}_t$. We suppose that there exists ${\bar{K}}_{i+1} \ge 0$ such that

$$\begin{aligned} V^{(N)}(t_{i+1},y_{t_{i+1}}) \le {\bar{K}}_{i+1}(1+\Vert y_{t_{i+1}}\Vert _\infty ^2), \quad \forall y_{t_{i+1}} \in {\mathbf {X}}_{t_{i+1}}. \end{aligned}$$

Taking $a=0$, we have from (3.2)

$$\begin{aligned} V^{(N)}(t,x_t)&\le K_l(1+\Vert x_t\Vert _\infty ^2) \varDelta _N +{\bar{K}}_{i+1} \left( 1+\sup _{b \in B}\Vert \xi _{t_{i+1}}^{(t,x_t),0,b} \Vert _\infty ^2 \right) . \end{aligned}$$

Thus, we have

$$\begin{aligned} V^{(N)}(t,x_t)&\le \left[ K_l\varDelta _N +{\bar{K}}_{i+1} \left\{ 1+2K_f(2+K_f)\varDelta _N \right\} \right] \\&\quad \times (1+\Vert x_t\Vert _\infty ^2) =:{\bar{K}}_i (1+\Vert x_t\Vert _\infty ^2), \end{aligned}$$

where we suppose $\varDelta _N \le 1$. Recalling (A4), we have the recursive equation of ${\bar{K}}_i$:

$$\begin{aligned}&{\bar{K}}_N=K_\varPhi , \\&{\bar{K}}_{i}= K_l\varDelta _N +{\bar{K}}_{i+1} \left[ 1+2K_f(2+K_f)\varDelta _N \right] \quad (i=N-1,N-2,\ldots ,0). \end{aligned}$$

Solving the above recursive equation, we have

$$\begin{aligned} {\bar{K}}_i&\le -\frac{K_l}{2K_f(2+K_f)} +\{ 1+ 2K_f (2+K_f) \varDelta _N \}^{N} \left( K_\varPhi +\frac{K_l}{2K_f(2+K_f)} \right) . \end{aligned}$$

Noting that $1+x \le e^x$ for $x \ge 0$, we have

$$\begin{aligned} {\bar{K}}_i \le -\frac{K_l}{2K_f(2+K_f)} +e^{ 2K_f (2+K_f) T} \left( K_\varPhi +\frac{K_l}{2K_f(2+K_f)} \right) . \end{aligned}$$

$\square$

Lemma A.1

Let $0<\epsilon <1,$ $t_i \le t < t_{i+1}$ and $x_t \in {\mathbf {X}}_{t}$. For given $b, b_{k} \in B \ (k=i+1,i+2,\ldots ,N-1),$ let $a^{\epsilon }, a^{\epsilon }_{k} \in A \ (k=i+1,i+2,\ldots , N-1)$ be given recursively in the following way:

$$\begin{aligned}&\sup _{b \in B}(T_{t,t_{i+1}}^{a^\epsilon ,b}V^{(N)}(t_{i+1},\cdot ))(x_t) < V^{(N)}(t,x_t)+\epsilon (t_{i+1}-t), \end{aligned}$$

(A.1)

$$\begin{aligned}&\sup _{b \in B} (T_{t_k, t_{k+1}}^{a^\epsilon _{k},b} V^{(N)}(t_{k+1},\cdot ))(x^\epsilon _{t_k}) < V^{(N)}(t_k, x^\epsilon _{t_k}) +\epsilon \varDelta _N, \quad \ k=i+1,\ldots , N-1, \nonumber \\ \end{aligned}$$

(A.2)

where $x^\epsilon _{t_k} \ (k=i+1,i+2,\ldots ,N)$ are given by

$$\begin{aligned} x^\epsilon _{t_{i+1}}=\xi ^{(t,x_t),a^\epsilon ,b}_{t_{i+1}}, \quad x^\epsilon _{t_{k+1}} =\xi ^{(t_{k},x^\epsilon _{t_{k}}),a^\epsilon _{k},b_{k}}_{t_{k+1}} \quad (k=i+1,i+2,\ldots , N-1). \end{aligned}$$

(A.3)

Then there exists $C>0$ depending only on $K_f, K_\varPhi , K_l, T, \nu$ such that

$$\begin{aligned} |a^\epsilon |^2 (t_{i+1}-t)+\sum _{k=i+1}^{N-1}|a^\epsilon _{k}|^2\varDelta _N \le C(1+\Vert x_t\Vert _\infty ^2). \end{aligned}$$

Proof

By (A.1) and (A.2), we have

$$\begin{aligned}&l(t,x_t, a^\epsilon ,b)(t_{i+1}-t)+V^{(N)}(t_{i+1}, x^\epsilon _{t_{i+1}})<V^{(N)}(t,x_t)+\epsilon (t_{i+1}-t), \\&l(t_{k},x^\epsilon _{t_{k}},a^\epsilon _{k},b_{k})\varDelta _N +V^{(N)}(t_{k+1},x^\epsilon _{t_{k+1}}) <V^{(N)}(t_{k},x^\epsilon _{t_{k}}) +\epsilon \varDelta _N \\&(k=i+1,\ldots , N-1). \end{aligned}$$

Adding the above inequalities, we have

$$\begin{aligned}&l(t,x_t, a^\epsilon ,b)(t_{i+1}-t) +\sum _{k=i+1}^{N-1} l(t_{k},x^\epsilon _{t_{k}},a^\epsilon _{k},b_{k})\varDelta _N +\varPhi (x^\epsilon _{t_{N}}) \\&\quad <V^{(N)}(t,x_t)+\epsilon (T-t). \end{aligned}$$

Using Proposition 3.3, (A3) and (A4), we have

$$\begin{aligned} |a^\epsilon |^2(t_{i+1}-t) +\sum _{k=i+1}^{N-1}|a^\epsilon _{k}|^2\varDelta _N \le (2/\nu )\{ K(1+\Vert x_t\Vert _\infty ^2)+K_lT+K_\varPhi +T\}. \end{aligned}$$

$\square$

Proof of Proposition 3.4

If $t=T$, (3.7) holds with $L=L_\varPhi$ because of (A4). Consider the case where $t_i \le t< t_{i+1}$. We may suppose that $V^{(N)}(t,y_t) \ge V^{(N)}(t,x_t)$ without loss of generality. Let $0<\epsilon <1$. Let $k=i,i+1,\ldots , N-1$ and $x^\epsilon _{t_k}$ and $y^{\epsilon }_{t_k}$ be given. Here, we abuse the notation $t=t_i$. Take $a^\epsilon _{k} \in A$ and then $b^\epsilon _k \in B$ such that

$$\begin{aligned}&\sup _{b \in B}(T_{t_k,t_{k+1}}^{a^\epsilon _{k},b}V^{(N)}(t_{k+1}, \cdot ))(x^\epsilon _{t_k})<V^{(N)}(t_k,x^\epsilon _{t_k}) +\epsilon \varDelta _N, \\&V^{(N)}(t_k, y^\epsilon _{t_k})< (T_{t_k, t_{k+1}}^{a^\epsilon _{k},b^\epsilon _k}V^{(N)}(t_{k+1}, \cdot ))(y^\epsilon _{t_k}) +\epsilon \varDelta _N. \end{aligned}$$

Then, we have

$$\begin{aligned} V^{(N)}(t_k,y^\epsilon _{t_k})-V^{(N)}(t_k,x^\epsilon _{t_k})&\le (l(t_k,y^\epsilon _{t_k}, a^\epsilon _k,b^\epsilon _k) -l(t_k,x^\epsilon _{t_k}, a^\epsilon _k,b^\epsilon _k))\varDelta _N \nonumber \\&\quad +V^{(N)}(t_{k+1},y^\epsilon _{t_{k+1}})-V^{(N)}(t_{k+1},x^\epsilon _{t_{k+1}})\nonumber \\&\quad +2\epsilon (t_{k+1}-t_k), \end{aligned}$$

(A.4)

where $x^\epsilon _{t_{k+1}} =\xi ^{(t_k, x^\epsilon _{t_k}),a^\epsilon _{k},b^\epsilon _{k}}_{t_{k+1}}, \ y^\epsilon _{t_{k+1}} =\xi ^{(t_k, y^\epsilon _{t_k}),a^\epsilon _{k},b^\epsilon _{k}}_{t_{k+1}}.$ Thus we can obtain

$$\begin{aligned} V^{(N)}(t,y_t)-V^{(N)}(t,x_t)&\le \sum _{k=i}^{N-1} (l(t_{k}, y^\epsilon _{t_{k}},a^\epsilon _{k},b^\epsilon _{k}) -l(t_{k}, x^\epsilon _{t_{k}},a^\epsilon _{k},b^\epsilon _{k})) (t_{k+1}-t_{k}) \nonumber \\&\quad +\varPhi (y^\epsilon _{t_N})-\varPhi (x^\epsilon _{t_N}) +2\epsilon T. \end{aligned}$$

(A.5)

We will show that there exists $C>0$ such that for $k=i,i+1,\ldots , N$,

$$\begin{aligned} \Vert x^\epsilon _{t_k}\Vert _\infty \le C(1+\Vert x_t\Vert _\infty +\Vert y_t\Vert _\infty ), \quad \Vert y^\epsilon _{t_k}\Vert _\infty \le C(1+\Vert x_t\Vert _\infty +\Vert y_t\Vert _\infty ). \end{aligned}$$

(A.6)

We denote by C and $C_j$ ($j=1,2,\ldots$) constants depending on $L_f, K_f, L_l, K_l, \nu , L_\varPhi , K_\varPhi , T$ for the rest of the present proof. We will prove only the estimate of $\Vert y^\epsilon _{t_k}\Vert _\infty$ in (A.6) because that of $\Vert x^\epsilon _{t_k}\Vert _\infty$ can be proved similarly. By using (A2), we have

$$\begin{aligned} \Vert y^\epsilon _{t_{j+1}} \Vert _\infty&\le p \Vert y^\epsilon _{t_j} \Vert _\infty +r_{j} \quad (j=i,i+1,\ldots ,k-1), \end{aligned}$$

(A.7)

where

$$\begin{aligned} p=1+K_f\varDelta _N, \quad r_{j}=K_f(t_{j+1}-t_j)+K_f |a^\epsilon _{j}|(t_{j+1}-t_j), \end{aligned}$$

Using (A.7) recursively, we have

$$\begin{aligned} \Vert y^\epsilon _{t_k} \Vert _\infty&\le p^{k-i} \Vert y_{t}\Vert _\infty +\sum _{j=i}^{k-1}p^{k-1-j}r_j \le p^{k-i} \Vert y_{t}\Vert _\infty +p^{k-i-1}\sum _{j=i}^{k-1}r_{j}. \end{aligned}$$

Here, we use $p \ge 1$ in the last inequality. Since $p^N=(1+K_f\varDelta _N)^N \le e^{K_fT}$ because $1+x \le e^x$ for $x \ge 0$, we have

$$\begin{aligned} \Vert y^\epsilon _{t_k}\Vert _\infty&\le e^{K_fT} \Vert y_{t}\Vert _\infty +e^{K_fT}K_f T +e^{K_fT}K_f \sum _{j=i}^{N-1} |a^\epsilon _{j}| (t_{j+1}-t_j). \end{aligned}$$

(A.8)

By Cauchy–Schwarz inequality and Lemma A.1, we have

$$\begin{aligned} \sum _{j=i}^{N-1} |a^\epsilon _{l}| (t_{j+1}-t_j) \le \sqrt{T}\left\{ \sum _{j=i}^{N-1}|a^\epsilon _{l}|^2 (t_{j+1}-t_j) \right\} ^{1/2} \le C_1(1+\Vert x_t\Vert _\infty ). \qquad \end{aligned}$$

(A.9)

Hence we obtain the second inequality of (A.6).

Estimating the RHS of (A.5) with (A.6), we have

$$\begin{aligned}&\text {RHS of (A.5)} \nonumber \\&\quad \le {C}_2(1+\Vert x_t\Vert _\infty +\Vert y_t\Vert _\infty ) \left\{ \sum _{k=i}^{N-1} \Vert x^\epsilon _{t_k}-y^\epsilon _{t_k}\Vert _\infty (t_{k+1}-t_k) + \Vert x^\epsilon _{t_N}-y^\epsilon _{t_N}\Vert _\infty \right\} \nonumber \\&\qquad +C_2 \sum _{k=i}^{N-1}|a^\epsilon _{k}|\Vert x^\epsilon _{t_k} -y^\epsilon _{t_k}\Vert _\infty (t_{k+1}-t_k) +2\epsilon T. \end{aligned}$$

(A.10)

Noting that

$$\begin{aligned} \Vert x^\epsilon _{t_k}-y^\epsilon _{t_k}\Vert _\infty \le (1+L_f \varDelta _N)\Vert x^\epsilon _{t_{k-1}}-y^\epsilon _{t_{k-1}}\Vert _\infty \le (1+L_f\varDelta _N)^{k-i} \Vert x_t -y_t\Vert _\infty \end{aligned}$$

and $(1+L_f \varDelta _N)^N \le e^{L_fT}$, we have

$$\begin{aligned}&\text {RHS of (A.10)} \le C_3 \Vert x_t -y_t\Vert _\infty \\&\times \left\{ 1+\Vert x_t\Vert _\infty +\Vert y_t\Vert _\infty +\sum _{k=i}^{N-1}|a^\epsilon _{k}| (t_{k+1}-t_k) \right\} +2\epsilon T. \end{aligned}$$

By (A.9), we have

$$\begin{aligned} V^{(N)}(t,y_t)-V^{(N)}(t,x_t) \le C_4(1+\Vert x_t\Vert _\infty +\Vert y_t\Vert _\infty )\Vert x_t-y_t\Vert _\infty +2\epsilon T. \end{aligned}$$

Letting $\epsilon \rightarrow 0$, we obtain

$$\begin{aligned} V^{(N)}(t,y_t)-V^{(N)}(t,x_t) \le {C}_4(1+\Vert x_t\Vert _\infty +\Vert y_t\Vert _\infty )\Vert x_t-y_t\Vert _\infty . \end{aligned}$$

$\square$

Proof of Proposition 3.5

Note that we can use Lemma 4.3 for $\varphi =V^{(N)}$ because Proposition 3.4 holds. We suppose N is sufficiently large such that Lemma 4.3 holds for $\varphi =V^{(N)}$. More precisely, suppose $h_L > T/N$. We denote by $C_i$ $(i=1,2,\ldots )$ constants depending on only $L_f$, $M_f$, $K_f$, $L_l$, $M_l$, $K_l$, $\nu$, $L_\varPhi$, $K_\varPhi$, and T in the present proof.

Suppose $t_i \le t< s < t_{i+1}$ and $x_t \in {\mathbf {X}}_t$. We consider the case where $V^{(N)}(s,x_s(\cdot \wedge t)) \ge V^{(N)}(t,x_t)$. We omit the proof for the case where $V^{(N)}(s,x_s(\cdot \wedge t)) \le V^{(N)}(t,x_t)$ because it can be proved in a manner similar to the above case. Take ${a}^*=a^*(t,x_t;t_{i+1}-t)$ such that

$$\begin{aligned} {a}^*\in \mathop {{\text {arg} \, \text {min} }}\limits _{a \in A} \max _{b \in B} \{ l(t,x_t,a,b)(t_{i+1}-t)+V^{(N)}(t_{i+1},\xi ^{(t,x_t),a,b}_{t_{i+1}})\}. \end{aligned}$$

(A.11)

We note that from Lemma 4.3 with Proposition 3.4

$$\begin{aligned} |{a}^*| \le C_1 (1+\Vert x_t\Vert _\infty ), \end{aligned}$$

(A.12)

where $C_1={\hat{C}}_L$ with L of Proposition 3.4 which depends on $K_f$, $K_l$, $K_\varPhi$, $L_f$, $L_l$, $L_\varPhi$, $\nu$, T. Using the definitions of $V^{(N)}(s,x_s(\cdot \wedge t))$ and $V^{(N)}(t,x_t)$, we have

$$\begin{aligned}&V^{(N)}(s,x_s(\cdot \wedge t))-V^{(N)}(t,x_t) \nonumber \\&\quad \le \sup _{b \in B} \{ |l(t,x_t,{a}^*,b)|(s-t) +|l(s,x_s(\cdot \wedge t),{a}^*,b)-l(t,x_t,{a}^*,b)|(t_{i+1}-s) \nonumber \\&\qquad +|V^{(N)}(t_{i+1},\xi ^{(s,x_s(\cdot \wedge t)),{a}^*,b}_{t_{i+1}}) -V^{(N)}(t_{i+1},\xi ^{(t,x_t),{a}^*,b}_{t_{i+1}})| \}. \end{aligned}$$

(A.13)

By (A3) and (A.12), we have

$$\begin{aligned} |l(t,x_t,{a}^*,b)|(s-t) \le C_2(1+\Vert x_t\Vert _\infty ^2)(s-t). \end{aligned}$$

(A.14)

Similarly, by (A3) and (A.12), we have

$$\begin{aligned}&|l(s,x_s(\cdot \wedge t),{a}^*,b)-l(t,x_t,{a}^*,b)| (t_{i+1}-s) \nonumber \\&\quad \le C_3(1+\Vert x_t\Vert _\infty ^2)(t_{i+1}-s)\omega ^{t,x_t}_l(s-t) . \end{aligned}$$

(A.15)

Using Proposition 3.4, we have

$$\begin{aligned}&|V^{(N)}(t_{i+1},\xi ^{(s,x_s(\cdot \wedge t)),{a}^*,b}_{t_{i+1}}) -V^{(N)}(t_{i+1},\xi ^{(t,x_t),{a}^*,b}_{t_{i+1}})| \\&\quad \le L(1+\Vert \xi ^{(s,x_s(\cdot \wedge t)),{a}^*,b}_{t_{i+1}}\Vert _\infty +\Vert \xi ^{(t,x_t),{a}^*,b}_{t_{i+1}} \Vert _\infty ) \Vert \xi ^{(s,x_s(\cdot \wedge t)),{a}^*,b}_{t_{i+1}} -\xi ^{(t,x_t),{a}^*,b}_{t_{i+1}}\Vert _\infty . \end{aligned}$$

By the definitions of $\xi ^{(s,x_s(\cdot \wedge t)),{a}^*,b}_{t_{i+1}}$ and $\xi ^{(t,x_t),{a}^*,b}_{t_{i+1}}$ with (A2) and (A.12), we have

$$\begin{aligned} \Vert \xi ^{(s,x_s(\cdot \wedge t)),{a}^*,b}_{t_{i+1}}\Vert _\infty \le C_4(1+\Vert x_t\Vert _\infty ), \quad \Vert \xi ^{(t,x_t),{a}^*,b}_{t_{i+1}} \Vert _\infty \le C_4(1+\Vert x_t\Vert _\infty ). \end{aligned}$$

Using the definitions of $\xi ^{(s,x_s(\cdot \wedge t)),{a}^*,b}_{t_{i+1}}$ and $\xi ^{(t,x_t),{a}^*,b}_{t_{i+1}}$ with (A2) and (A.12), we have

$$\begin{aligned} \Vert \xi ^{(s,x_s(\cdot \wedge t)),{a}^*,b}_{t_{i+1}} -\xi ^{(t,x_t),{a}^*,b}_{t_{i+1}}\Vert _\infty \le C_5(1+\Vert x_t\Vert _\infty ) \{ (s-t) +(t_{i+1}-s)\omega ^{t,x_t}_f(s-t) \}. \end{aligned}$$

Thus, there exists a constant $C_6>0$ such that

$$\begin{aligned} V^{(N)}(s,x_s(\cdot \wedge t))-V^{(N)}(t,x_t)\le & {} C_6 (1+\Vert x_t\Vert _\infty ^2) \{ (s-t) \nonumber \\&\quad +(t_{i+1}-s)\omega ^{t,x_t}(s-t) \}, \end{aligned}$$

(A.16)

where $\omega ^{t,x_t}=\omega ^{t,x_t}_f+\omega ^{t,x_t}_l$.

We suppose $t_i \le t <s=t_{i+1}$. Since the arguments to obtain (A.16) are valid for $s=t_{i+1}$, we have

$$\begin{aligned} V^{(N)}(t_{i+1},x_{t_{i+1}}(\cdot \wedge t))-V^{(N)}(t,x_t) \le C_6 (1+\Vert x_t\Vert _\infty ^2) (t_{i+1}-t). \end{aligned}$$

From the definition of $V^{(N)}(t,x_t)$, we have

$$\begin{aligned} V^{(N)}(t,x_t) =\inf _{a \in A} \sup _{b \in B} \{ l(t,x_t,a,b)(t_{i+1}-t) +V^{(N)}(t_{i+1},\xi _{t_{i+1}}^{(t,x_t),a,b}) \} \end{aligned}$$

from which we have

$$\begin{aligned}&V^{(N)}(t,x_t)-V^{(N)}(t_{i+1},x_{t_{i+1}}(\cdot \wedge t)) \nonumber \\&\quad =\inf _{a \in A} \sup _{b \in B} \{ l(t,x_t,a,b)(t_{i+1}-t) +V^{(N)}(t_{i+1},\xi _{t_{i+1}}^{(t,x_t),a,b})\nonumber \\&\qquad -V^{(N)}(t_{i+1},x_{t_{i+1}}(\cdot \wedge t)) \}. \end{aligned}$$

(A.17)

If we take $a=0$, we have

$$\begin{aligned}&V^{(N)}(t,x_t)-V^{(N)}(t_{i+1},x_{t_{i+1}}(\cdot \wedge t)) \\&\quad \le \sup _{b \in B} \{ l(t,x_t,0,b)(t_{i+1}-t) +V^{(N)}(t_{i+1},\xi _{t_{i+1}}^{(t,x_t),0,b}) -V^{(N)}(t_{i+1},x_{t_{i+1}}(\cdot \wedge t)) \} \\&\quad \le \sup _{b \in B}|l(t,x_t,0,b)|(t_{i+1}-t) +\sup _{b \in B} |V^{(N)}(t_{i+1},\xi _{t_{i+1}}^{(t,x_t),0,b})\\&\qquad -V^{(N)}(t_{i+1},x_{t_{i+1}}(\cdot \wedge t))|. \end{aligned}$$

By (A3), we have

$$\begin{aligned} \sup _{b \in B}|l(t,x_t,0,b)|(t_{i+1}-t) \le C_7 (1+\Vert x_t\Vert _\infty ^2) (t_{i+1}-t). \end{aligned}$$

Using Proposition 3.4 and (A2), we have

$$\begin{aligned}&|V^{(N)}(t_{i+1},\xi _{t_{i+1}}^{(t,x_t),0,b}) -V^{(N)}(t_{i+1},x_{t_{i+1}}(\cdot \wedge t))| \\&\quad \le L(1+\Vert x_t\Vert _\infty +\Vert \xi _{t_{i+1}}^{(t,x_t),0,b}\Vert _\infty ) \Vert x_{t_{i+1}}(\cdot \wedge t)-\xi _{t_{i+1}}^{(t,x_t),0,b}\Vert _\infty \\&\quad \le C_8 (1+\Vert x_t\Vert _\infty ^2)(t_{i+1}-t). \end{aligned}$$

Thus, we obtain

$$\begin{aligned} V^{(N)}(t,x_t)-V^{(N)}(t_{i+1},x_{t_{i+1}}(\cdot \wedge t)) \le (C_7+C_8)(1+\Vert x_t\Vert _\infty ^2)(t_{i+1}-t). \end{aligned}$$

Hence, we have

$$\begin{aligned} |V^{(N)}(t_{i+1},x_{t_{i+1}}(\cdot \wedge t))-V^{(N)}(t,x_t)| \le C_9(1+\Vert x_t\Vert _\infty ^2)(t_{i+1}-t). \end{aligned}$$

Suppose $0 \le t_i \le t<t_{i+1} \le t_j < s \le t_{j+1}$. Note that

$$\begin{aligned}&|V^{(N)}(s,x_s(\cdot \wedge t))-V^{(N)}(t,x_t)| \nonumber \\&\quad \le |V^{(N)}(s,x_s(\cdot \wedge t))-V^{(N)}(t_j, x_{t_j}(\cdot \wedge t))| \nonumber \\&\qquad +\sum _{k=i+1}^{j-1}|V^{(N)}(t_{k+1},x_{t_{k+1}}(\cdot \wedge t)) -V^{(N)}(t_{k},x_{t_{k}}(\cdot \wedge t))| \nonumber \\&\qquad +|V^{(N)}(t_{i+1},x_{t_{i+1}}(\cdot \wedge t))-V^{(N)}(t,x_t)|. \end{aligned}$$

(A.18)

Let $t_j<s<t_{j+1}$. Applying the first case of (3.8) (resp. the second case of (3.8)) to the first term of (A.18) (resp. the second and third terms of (A.18)), we have

$$\begin{aligned} \text {RHS of (A.18)}&\le C_{10}(1+\Vert x_{t}\Vert _\infty ^2) \{(s-t_j)+(t_{j+1}-s)\omega ^{t,x_t}(s-t_j)\} \\&\quad +\sum _{k=i+1}^{j-1} C_{10}(1+\Vert x_t\Vert _\infty ^2)(t_{k+1}-t_{k}) +C_{10}(1+\Vert x_t\Vert _\infty ^2)(t_{i+1}-t) \\&=C_{10}(1+\Vert x_{t}\Vert _\infty ^2)\{ (s-t)+(t_{j+1}-s)\omega ^{t,x_t}(s-t_j) \}. \end{aligned}$$

Let $t_j<s =t_{j+1}$. Applying the second case of (3.8) to each term of (A.18), we obtain

$$\begin{aligned} \text {RHS of (A.18)}&\le C_{11}(1+\Vert x_{t}\Vert _\infty ^2) (s-t_j) \\&\quad +\sum _{k=i+1}^{j-1} C_{11}(1+\Vert x_t\Vert _\infty ^2)(t_{k+1}-t_{k}) +C_{11}(1+\Vert x_t\Vert _\infty ^2)(t_{i+1}-t) \\&=C_{11}(1+\Vert x_{t}\Vert _\infty ^2)(s-t). \end{aligned}$$

$\square$